and the lowering of the surface is
| p − p0 | − z = k log | ρ | − z = −k log ( 1 − | z | ) − z ≈ | z2 |
| ρ0 | ρ0 | k | 2k |
(20)
as before in (17).
16. Centre of Pressure.—A plane area exposed to fluid pressure on one side experiences a single resultant thrust, the integrated pressure over the area, acting through a definite point called the centre of pressure (C.P.) of the area.
Thus if the plane is normal to Oz, the resultant thrust
R = ∫ ∫ p dx dy,
(1)
and the coordinates x, y of the C.P. are given by
xR = ∫ ∫ xp dx dy, yR = ∫ ∫ yp dx dy.